Let the solution curve x = x ( y ) , 0 < y < π 2 , of the differential equation log e cos y 2 cos y d x - 1 + 3 x log e cos y sin y d y = 0 satisfy x π 3 = 1 2 log e 2 . If x π 6 = 1 log e m - log e n , where m and n are coprime, then m n is equal to

Step-by-step solution

Given, log e cos y 2 cos y   d x - 1 + 3 x log e cos y sin y   d y = 0 ⇒ d x d y + - 3 tan y lncos y x = sin y In cos y 2 · cos y Which is a linear differential equation, So, Integrating factor will be, I F = e 3 ∫ - tan y lncos y d y = lncos y 3 So, solution of differential equation is given by, x lncos y 3 = ∫ sin y cos y lncos y d y ⇒ x lncos y 3 = - lncos y 2 2 + c Now using the given value of x π 3 = 1 2 ln 2 We get, c = 0 Hence, x = - 1 2 ln cos   y Now finding